Question

how do you find the x-intercepts and the vertex of this parabola: y= -x²-5x ?

Answer 1

$-x^2-5x=0\\ -x(x+5)=0\\ x=0 \vee x=-5$

The x-intercepts are $(0,0)$ and $(-5,0)$.

$\hbox{vertex}=(h,k)\\ h=\frac{x_1+x_2}{2},k=y(h)\\\\ h=\frac{0-5}{2}=\frac{-5}{2}=-2.5\\ k=y(-2.5)=-(-2.5)^2-5\cdot(-2.5)\\ k=-6.25+12.5\\ k=6.25\\\\ \hbox{vertex}=(2.5,6.25)$